使用zoo::na.fill(0)可以通過填寫0所有的NA，而這將改變變數屬性，例如，變數的屬性amount是num在資料幀的MD，之后zoo::na.fill(0)的屬性變化chr。有什么方法可以像這樣簡單地填充 nana.fill并保持變數屬性？
md <- data.frame(cat=c('a','b','d',NA,'E',NA),
subcat=c('A','C',NA,NA,NA,'D'),
amount=c(1,2,NA,5,NA,8))
md %>% zoo::na.fill(0)
uj5u.com熱心網友回復：
這是一個不使用動物園的解決方案。
library(dplyr)
md2 <- md %>%
mutate(across(where(is.factor), as.character)) %>%
mutate(across(where(is.character), function(x) { replace(x, is.na(x), "0") } )) %>%
mutate(across(where(is.numeric), function(x) { replace(x, is.na(x), 0) } )) %>%
mutate(across(where(is.character), as.factor))
如果你愿意，你可以把它包裝成一個自定義函式，就像 zoo 的 fill-na 方法一樣容易地使用它，例如
FillNA <- function(df){
df2 <- df %>%
mutate(across(where(is.factor), as.character)) %>%
mutate(across(where(is.character), function(x) { replace(x, is.na(x), "0") } )) %>%
mutate(across(where(is.numeric), function(x) { replace(x, is.na(x), 0) } )) %>%
mutate(across(where(is.character), as.factor))
return(df2)
}
這里是型別的驗證：
> str(md)
'data.frame': 6 obs. of 3 variables:
$ cat : Factor w/ 4 levels "a","b","d","E": 1 2 3 NA 4 NA
$ subcat: Factor w/ 3 levels "A","C","D": 1 2 NA NA NA 3
$ amount: num 1 2 NA 5 NA 8
str(FillNA(md))
'data.frame': 6 obs. of 3 variables:
$ cat : Factor w/ 5 levels "0","a","b","d",..: 2 3 4 1 5 1
$ subcat: Factor w/ 4 levels "0","A","C","D": 2 3 1 1 1 4
$ amount: num 1 2 0 5 0 8
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